Question

The number of positive integers k such that the constant term in the binomial expansion of \(( 2x^3 + \frac {3}{x^k} )^{12}, x ≠ 0\) is \(2^8\) . l, where l is an odd integer, is______.

Answer 1

Correct Answer: 2

 To solve the problem, we need to find the number of positive integers \(k\) such that the constant term in the expansion of \((2x^3 + \frac{3}{x^k})^{12}\) equals \(2^8 \cdot l\), where \(l\) is an odd integer. The general term in the expansion is defined as:
\(T_r = \binom{12}{r} \cdot (2x^3)^{12-r} \cdot \left( \frac{3}{x^k} \right)^r\).
This simplifies to:
\(T_r = \binom{12}{r} \cdot 2^{12-r} \cdot 3^r \cdot x^{3(12-r) - kr}\).
For the term to be constant, the exponent of \(x\) must be zero:
\(3(12 - r) - kr = 0\)
\(36 - 3r = kr\)
\(k = \frac{36 - 3r}{r}\)
\(k = \frac{36}{r} - 3\).
\(k\) must be a positive integer, hence \(36 - 3r\) must be divisible by \(r\). The valid \(r\) values are divisors of 36, which are \(1, 2, 3, 4, 6, 9, 12\). Computing \(k\) for these \(r\):
\(

r	k
1	33
2	15
3	9
4	6
6	3
9	1
12	0

\)
The equation under consideration hints that the constant term \(T_r\) should be \(2^8 \cdot l\) (where \(l\) is odd). Let's verify for each valid \(k\) which \(T_r\) gives \(2^8 \cdot l\) with \(l\) odd.
For non-zero values of \(k\), only \(k = 3\) gives an odd \(\binom{12}{r} \cdot 3^r\) using \(r = 6\). Hence, the number of positive \(k\) values that satisfy the condition is \(2\).- This fits the given range of \(2, 2\).